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jekas [21]
3 years ago
6

Alex found a new fruit punch recipe that calls for cranberry juice and lemon-lime soda. If cranberry juice costs $3.60 per bottl

e and lemon-lime soda costs $1.40 per bottle and the recipe calls for 2 times as many bottles of lemon-lime soda as cranberry juice, at most how many bottles of cranberry juice can he buy if he only has $38.40
Mathematics
1 answer:
belka [17]3 years ago
6 0

Answer:

Alex can buy 12 bottles of lemon-lime soda and 6 bottles of cranberry juice.

Step-by-step explanation:

Since Alex found a new fruit punch recipe that calls for cranberry juice and lemon-lime soda, if cranberry juice costs $ 3.60 per bottle and lemon-lime soda costs $ 1.40 per bottle and the recipe calls for 2 times as many bottles of lemon-lime soda as cranberry juice, to determine at most how many bottles of cranberry juice can he buy if he only has $ 38.40 the following calculation must be performed:

20 x 1.40 + 10 x 3.60 = 64

10 x 1.40 + 5 x 3.60 = 32

12 x 1.40 + 6 x 3.60 = 38.40

Therefore, Alex can buy 12 bottles of lemon-lime soda and 6 bottles of cranberry juice.

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Please help with this problem
Andre45 [30]

Answer:

The length of the short side is 14.5 units, the length of the other short side is 18.5 units, and the length of the longest side is 23.5 units.

Step-by-step explanation:

The Pythagorean Theorem

<em> If a and b are the lengths of the legs of a right triangle and c is the length of the hypotenuse, then the sum of the squares of the lengths of the legs is equal to the square of the length of the hypotenuse. </em>

This relationship is represented by the formula:

                                                     a^2+b^2=c^2

Applying the Pythagorean Theorem  to find the lengths of the three sides we get:

(x)^2+(x+4)^2=(x+9)^2\\\\2x^2+8x+16=x^2+18x+81\\\\2x^2+8x-65=x^2+18x\\\\2x^2-10x-65=x^2\\\\x^2-10x-65=0

Solve with the quadratic formula

\mathrm{For\:a\:quadratic\:equation\:of\:the\:form\:}ax^2+bx+c=0\mathrm{\:the\:solutions\:are\:}

x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}

\mathrm{For\:}\quad a=1,\:b=-10,\:c=-65:\quad x_{1,\:2}=\frac{-\left(-10\right)\pm \sqrt{\left(-10\right)^2-4\cdot \:1\left(-65\right)}}{2\cdot \:1}\\\\x_{1}=\frac{-\left(-10\right)+ \sqrt{\left(-10\right)^2-4\cdot \:1\left(-65\right)}}{2\cdot \:1}=5+3\sqrt{10}\\\\x_{2}=\frac{-\left(-10\right)- \sqrt{\left(-10\right)^2-4\cdot \:1\left(-65\right)}}{2\cdot \:1}=5-3\sqrt{10}

Because a length can only be positive, the only solution is

x=5+3\sqrt{10}\approx 14.5

The length of the short side is 14.5, the length of the other short side is 14.5+4=18.5, and the length of the longest side is 14.5+9=23.5.

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Answer:

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Step-by-step explanation:

y=cd-7

y+7=c×d

y+7/d=c

c=y+7/d

:)

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